Re: Killer Sudoku 21 May 2021

That puzzle (link will expire) does seem quite difficult. While I can't offer a full analytical solution, here are some of the observations that helped me solve it. (I

*did* use "pencil" numbers for this one

)

(A) The sum of the numbers in cells **f2** and **f3** is **5**, so those two cells must contain either [1,4] or [2,3]. (In the former case, 4 must be in f3 because the 10 cage in that quadrant can't contain [1,4,5] while 1 is also in f3.)

(B) In column **d**, the sum of the numbers in the four cells outside the 15 and 19 cages is 45 - 15 - 19 = **11**, which means that those cells must contain [1,2,3,5].

(C) If the **10** cage in cells **c3**, **c4** and **d4** contains the number **1**, it must be in **d4** (because the other cells are "covered" by the 1 in **c5**). If not, the numbers in that cage must be [2,3,5].

(D) In the **upper left quadrant**, the sum of all the cages that are wholly or partially in the quadrant is 10 + 11 + 12 + 13 + 14 = 60, of which 45 is inside the quadrant and **15** in the combined four "outie" cells of the cages 10, 13 and 14 (cells **a4**, **c4**, **d4** and **d1**, of which the latter two are among the four cells mentioned in (B).

(E) I tried to determine where **the number 9** could go within the same quadrant. Its upper row is "covered" by the 9 in **i1**, and **c3** obviously can't contain 9 either. When I placed 9 in **a2** or **c2**, there was no correct way to complete rows 1 and 2. That left the **11** cage (cells **b2** and **b3**) and cell **a3** as the only possible positions of 9 within that quadrant.

The latter two observations made me think that the rather small sum of the four "outies" combined with the knowledge that

**a4** had to contain a comparatively high number (

**5**,

**6**,

**8** or

**9**) might be enough of a constraint to permit some inferences. I tried putting

**9** in that cell, which soon turned out not to permit any correct solution to columns

**c** and

**d** (the

**21** cage at the bottom).

Keeping my focus on the number

**9**, I then tried putting it in

**a3**, with

**5** in

**a4**. Apropos (C), I don't remember if I managed to prove (using columns c and d) that the

**10** cage couldn't contain

**1** under this hypothesis, or if [2,3,5] in that cage was just a sub-hypothesis I was checking, but at any rate it led to a correct solution to the puzzle.

Even assuming that [2,3,5] in the

**10** cage was the result of a valid inference, I would of course still have to check and exclude both variants of [6,8] in the

**14** cage in cells

**a3** and

**a4** (with [2,9] in the

**11** cage within the upper left quadrant) to make sure that the solution I'd found was unique.

Working through the branches of trial and error this way after you've found a correct solution may give you a better understanding of the set of conditions/constraints that force the unique solution to the puzzle. You may find (or at least approach) a full analytical solution

instead of a

gamebook-style catalog of all the branches (and twigs) of trial and error where all paths execpt one lead to a dead end

But I'm trying to cut back on the puzzling so I won't attempt any further analysis of this one